sin6 theta + cos6 theta = 1 - 3 sin2 theta cos2 theta
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Answer:
Step-by-step explanation:
Sin⁶θ + Cos⁶θ
= (Sin²θ)³ + (Cos²θ)³
// We know that (a + b)³ = a³ + b³ + 3ab(a + b) => a³ + b³ = (a + b)³ - 3ab(a + b)
= (Sin²θ + Cos²θ)³ - 3Sin²θCos²θ(Sin²θ + Cos²θ)
= 1 - 3Sin²θCos²θ. (∵ Sin²θ + Cos²θ = 1)
= R.H.S
Hence proved.
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